Let f be defined on the natural numbers as follow: f(1)=1 and for `n gt 1, f(n)=f[f(n-1)]+f[n-f(n-1)]`, the value of `1/(30)sum_(r=1)^(20)f(r)` is

Answer

To solve the problem, we need to find the function \( f(n) \) defined recursively and then compute the sum \( \frac{1}{30} \sum_{r=1}^{20} f(r) \).

### Step 1: Define the function \( f(n) \)

We know:
- \( f(1) = 1 \)
- For \( n > 1 \), \( f(n) = f(f(n-1)) + f(n - f(n-1)) \)

### Step 2: Calculate values of \( f(n) \) for \( n = 2, 3, \ldots, 20 \)

1. **Calculate \( f(2) \)**:
\[ f(2) = f(f(1)) + f(2 - f(1)) = f(1) + f(1) = 1 + 1 = 2 \]

2. **Calculate \( f(3) \)**:
\[ f(3) = f(f(2)) + f(3 - f(2)) = f(2) + f(3 - 2) = 2 + f(1) = 2 + 1 = 3 \]

3. **Calculate \( f(4) \)**:
\[ f(4) = f(f(3)) + f(4 - f(3)) = f(3) + f(4 - 3) = 3 + f(1) = 3 + 1 = 4 \]

Continuing this pattern, we can see that:
- \( f(5) = 5 \)
- \( f(6) = 6 \)
- \( f(7) = 7 \)
- \( f(8) = 8 \)
- \( f(9) = 9 \)
- \( f(10) = 10 \)
- \( f(11) = 11 \)
- \( f(12) = 12 \)
- \( f(13) = 13 \)
- \( f(14) = 14 \)
- \( f(15) = 15 \)
- \( f(16) = 16 \)
- \( f(17) = 17 \)
- \( f(18) = 18 \)
- \( f(19) = 19 \)
- \( f(20) = 20 \)

### Step 3: Calculate the sum \( \sum_{r=1}^{20} f(r) \)

Since we found that \( f(r) = r \) for \( r = 1, 2, \ldots, 20 \):
\[ \sum_{r=1}^{20} f(r) = 1 + 2 + 3 + \ldots + 20 \]

Using the formula for the sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
For \( n = 20 \):
\[ \sum_{r=1}^{20} r = \frac{20(20 + 1)}{2} = \frac{20 \times 21}{2} = 210 \]

### Step 4: Calculate \( \frac{1}{30} \sum_{r=1}^{20} f(r) \)

Now we compute:
\[ \frac{1}{30} \sum_{r=1}^{20} f(r) = \frac{1}{30} \times 210 = 7 \]

### Final Answer:
\[ \frac{1}{30} \sum_{r=1}^{20} f(r) = 7 \]